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Radiation Field) 221B Lecture Notes Quantum Field Theory III (Radiation Field) 1 Quantization of Radiation Field Early development of quantum mechanics was led by the fact that electro- magnetic radiation was quantized: photons. Now that we have gone through quantization of a classical field (Schr¨odingerfield so far), we can proceed to quantize the Maxwell field. The basic idea is pretty much the same, except that there are subtletites associated with the gauge invariance of the vector potential. 1.1 Classical Maxwell Field The vector potential A~ and the scalar potential φ are combined in the four- vector potential Aµ = (φ, A~): (1) Throughout the lecture notes, we use the convention that the metrix gµν = ν ~ diag(+1; 1; 1; 1) and hence Aµ = gµνA = (φ, A). The four-vector coordinate− is−xµ =− (ct; ~x), and correspondingly the four-vector− derivative is 1 @ ~ @µ = ( ; ). The field strength is defined as Fµν = @µAν @νAµ, and c @t r − ~_ ~ ~ hence F0i = @0Ai @iA0 = A=c φ = E, while Fij = @iAj @jAi = ~ j ~ i − ~ k − − r − iA + jA = ijkB . −r r − In the unit we have been using where the Coulomb potential is QQ0=r without a factor of 1=4π0, the action for the Maxwell field is 1 µν µ 1 ~ 2 ~ 2 µ S = Z dtd~x − F Fµν Aµj = Z dtd~x E B Aµj : (2) 8π − 8π − − I included a possible source term for the Maxwell field (electric current den- sity) jµ = (ρ,~j=c). For a point particle of charge e, the charge density is _ ρ = eδ(~x ~x0) while the current density is ~j = e~xδ(~x ~x0). They satisfy the current conversation− law − µ 1 @ ~ @µj = ρ + ~j! = 0: (3) c @t r · 1 The gauge invariance of the Maxwell field is that the vector potential Aµ and Aµ + @µ! (where ! is an arbitrary function of spacetime) give the same field strength and hence the same action. Using this invariance, one can always choose a particular gauge. For most purposes of non-relativistic systems encountered in atomic, molecular, condensed matter, nuclear and astrophysics, Coulomb gauge is the convenient choice, while for highly rel- ativistic systems such as in high-energy physics. We use Coulomb gauge in this lecture note: ~ A~ = 0: (4) r · A word of caution is that this gauge condition is not Lorentz-invariant, i.e., the gauge condition is not frame independent. Therefore, when you go to a different frame of reference, you also need to perform a gauge transformation at the same time to preserve the Coulomb gauge condition. Another point is that, in the Coulomb gauge, the Gauss' law is ~ E~ = 4πj0; (5) r · _ where j0 is the charge density. Because E~ = A~ ~ φ and the Coulomb gauge condition, we find the Poisson equation − − r ∆φ = 4πρ, (6) − and hence 1 φ(~x;t) = Z d~y ρ(~y; t): (7) ~x ~y j − j Note that the potential is not retarded, but instantaneous (i.e., determined by the charge distribution at the same instance). Hamiltonian for a particle of electric charge e in the presence of the Maxwell field is (~p e A~)2 H = − c + eφ. (8) 2m 1.2 Quantization In order to quantize the Maxwell field, we first determine the \canonically conjugate momentum" for the vector potential A~. Following the definition pi = @L=@q_i in particle mechanics, we define the canonically conjugate mo- mentum i @ 1 i 1 1 i 1 i π = L = E = A_ + ~ φ : (9) @A_ i −4πc 4π c2 c r 2 In the absence of extrernal sources, the scalar potential identically vanishes in Coulomb gauge (see Eq. (7)). We drop it entirely in this section.1 Following the normal canonical commutation relation [qi; pj] = ihδ¯ ij, we set up the commutation relation 1 [Ai(~x); πj(~y)] = [Ai(~x); A_ j(~y)] = ihδ¯ ijδ(~x ~y): (10) 4πc2 − To satisfy this commutation relation, we introduce the photon creation and annihilation operators i j ij [a (~p); a y(~q)] = δ δ~p;~q; (11) and expand the vector potential and its time derivative as 2πhc¯ 2 1 Ai(~x) = s (ai(~p)ei~p ~x=¯h + ai (~p)e i~p ~x=¯h) (12) 3 X · y − · L ~p p!p 2πhc¯ 2 A_ i(~x) = s ( i ! )(ai(~p)ei~p ~x=¯h ai (~p)e i~p ~x=¯h): (13) 3 X p p · y − · L ~p − − Here, !p = Ep=h¯ = c ~p =h¯ is the angular frequency for the photon. You can check that this momentum-modej j expansion together with the commutation relation Eq. (11) reproduces the canonical commutation relation Eq. (10) as follows. [Ai(~x); A_ j(~y)] 2πhc¯ 2 = ( i)[ai(~p)ei~p ~x=¯h + ai (~p)e i~p ~x=¯h; aj(~p)ei~q ~y=¯h aj (~q)e i~q ~y=¯h] 3 X · y − · · y − · L ~p;~q − − 2πhc¯ 2 = i δij(ei~p (~x ~y) + e i~p (~x ~y)) 3 X · − − · − L ~p = 4πhc¯ 2iδ(~x ~y): (14) − At the last step, we used the correspondence in the large volume limit = P~p L3 d~p=(2πh¯)3. R The problem with what we have done so far is that we have not imposed the Coulomb gauge condition Eq. (4) on the vector potential yet. Acting the divergence on the momentum-mode expansion Eq. (12), we need ~p ~a(~p) = 0: (15) · 1Even when we have matter particles or fields, their charge density operator commutes with the vector potential, and hence the discussion here goes through unmodified. 3 The meaning of this equation is obvious: there is no longitudinal photon. There are only two transverse polarizations. To satisfy this constraint while retaining the simple commutation relations among creation and annihilation operators, we introduce the polarization vectors. When ~p = (0; 0; p), the pos- itive helicity (right-handed) circular polarization has the polarization vector ~+ = (1; i; 0)=p2, while the negative helicity (left-handed) circular polariza- tion is represented by the polarization vector ~ = (1; i; 0)=p2. In general, for the momentum vector ~p = p(sin θ cos φ, sin −θ sin φ, cos− θ), the circular po- larization vectors are given by 1 (~p) = ( 1 + i2) ; (16) ± p2 ± where the linear polarization vectors are given by 1(~p) = (cos θ cos φ, cos θ sin φ, sin θ); (17) − 2(~p) = ( sin φ, cos φ, 0): (18) − Given the polarization vectors, we re-expand the vector potential in terms of the creation/annihilation operators 2 i s2πhc¯ 1 i i~p ~x=¯h i i~p ~x=¯h A (~x) = ( (~p)a (~p)e · + (~p)∗ay (~p)e− · ) (19) 3 X X ± ± L ~p p!p ± ± ± 2πhc¯ 2 _ i s i i~p ~x=¯h i i~p ~x=¯h A (~x) = ( ip!p) ( (~p)a (~p)e · (~p)∗ay (~p)e− · ): 3 X X ± ± L ~p − ± − ± ± (20) With this expansion, the Coulomb gauge condition is automatically satisfied, while the creation/annihilation operators obey the commutation relations [aλ(~p); ay (~q)] = δλ,λ δ~p;~q (21) λ0 0 for λ, λ0 = . We could also have used the linearly polarized photons ± 2 2 i s2πhc¯ 1 i i~p ~x=¯h i i~p ~x=¯h A (~x) = ( (~p)a (~p)e + (~p)ay (~p)e ) (22) 3 X X λ λ · λ λ − · L ~p p!p λ=1 2 2 i s2πhc¯ i i~p ~x=¯h i i~p ~x=¯h A_ (~x) = ( i ! ) ( (~p)a (~p)e (~p)ay (~p)e ): 3 X p p X λ λ · λ λ − · L ~p − λ=1 − (23) 4 with [aλ(~p); ay (~q)] = δλ,λ δ~p;~q (24) λ0 0 for λ, λ0 = 1; 2. Clearly, two sets of operators are related by 1 i a1 = (a+ + a ); a2 = (a+ a ): (25) p2 − p2 − − Once we have the mode expansion for the vector potential, one can work out the Hamiltonian 1 ~ 2 ~ 2 1 H = Z d~x E + B = h!¯ p(ay (~p)a (~p) + ): (26) X X ± ± 8π ~p 2 ± Becauseh! ¯ p = c ~p , the dispersion relation for the photon is the familiar one E = c ~p for a masslessj j relativistic particle. j j 2 Classical Electromagnetic Field Now that we found photons (particles) from the quantized radiation field, what is a classic electromagnetic field? To answer this question, we study the quantum-mechanical Hamiltonian for the photons in the presence of a source. Starting from the action Eq. (2), we find the Hamiltonian 1 2 2 1 H = Z d~x E~ + B~ A~ ~j 8π − c · = h!¯ ay (~p)a (~p) X p λ λ ~p,λ 2 1s2πhc¯ 1 ~ ~ y ! 3 (~λ(~p) j(~p)aλ(~p) + ~λ∗ (~p) j∗(~p)aλ(~p)) : (27) − c L p!p · · Here, we omitted the zero-point energy because it is not relevant for the discussions below, and the Fourier modes of the source is defined by i~p ~x=¯h ~j(~p) Z d~x~j(~x)e · ; (28) ≡ and ~j∗(~p) = ~j( ~p). The interesting point is that for each momentum ~p and polarization state− λ, the Hamiltonian Eq. (27) is of the type h!¯ (aya f ∗a ayf); (29) − − 5 whose ground state is h!¯ (aya f ∗a ayf) f = h!f¯ ∗f f : (30) − − j i − j i In other words, the ground state of the Hamiltonian for photons in the pres- ence of a source term is a coherent state f fλ(~p) , with j i ≡ Q~p,λ j i s2πh¯ 1 ~ fλ(~p) = 3 3=2~λ∗ j∗(~p): (31) L h!¯ p · The vector potential has an expectation value in the coherent state, given by f Ai(~x) f h j j i 2πhc¯ 2 1 = s (~i (~p)f (~p)ei~p ~x=¯h + ~i (~p)f (~p)e i~p ~x=¯h) X 3 2 λ λ · λ∗ λ∗ − · ~p,λ L h!¯ p 2πh¯ c = jk (~p)ei~p ~x=¯h i (~p)k (~p) + c:c:! : (32) 3 X 2 ∗ · X λ λ∗ L ~p h!¯ p λ Even though the expression is somewhat complicated, one can check that this expectation value satisfies 4π ~ f B~ f = ∆ f A~ f = ~j; (33) r × h j j i − h j j i c using the identity i j i j ij p p (~p) ∗(~p) = δ (34) X λ λ 2 λ − ~p together with the Coulomb gauge condition ~ A~ = 0.
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